High-Frequency Instabilities of the Ostrovsky Equation

Abstract

We study the spectral stability of small-amplitude periodic traveling waves of the Ostrovsky equation. We prove that these waves exhibit spectral instabilities arising from a collision of pair of non-zero eigenvalues on the imaginary axis when subjected to square-integrable perturbations on the whole real line. We also list all such collisions between pairs of eigenvalues on the imaginary axis and do a Krein signature analysis.

Appendix A: Lyapunov–Schmidt Procedure

Since

$$\begin{aligned} \partial _k F(w,c;k) := (2ck w'' + 4\beta k^3 w'''' -2k (w^2)'' + \gamma w) \quad \text {and} \quad \partial _c F(w,c;k) := k^2 w'' \end{aligned}$$

are continuous, $F:H^{4}({\mathbb {T}}) \times {\mathbb {R}} \times {\mathbb {R}}^+ \rightarrow L^2({\mathbb {T}})$ is $C^1$. Note that

$$\begin{aligned} L_0 e^{\pm iz}=0. \end{aligned}$$

For arbitrary $k>0$, we seek a non-trivial solution $w\in H^{4}({\mathbb {T}})$ near trivial solution $w\equiv 0$ of

$$\begin{aligned} F(w,c;k)=0 \end{aligned}$$

(A.1)

for some c near $c_0$, where $c_0$ is defined in (2.3). Let

$$\begin{aligned} w(z)=0+\frac{1}{2}ae^{iz}+\frac{1}{2}\bar{a}e^{-iz}+v(z) \quad \text {and}\quad c=c_0+r, \end{aligned}$$

where $a\in {\mathbb {C}}$ and $v\in H^{4}({\mathbb {T}})$ satisfying that

$$\begin{aligned} \int _{\mathbb {T}} v(z)\,e^{\pm iz}~dz=0, \end{aligned}$$

and $r\in {\mathbb {R}}$. Plugging these into (A.1) and using the fact that $L_0e^{\pm iz}=0$, we arrive at

$$\begin{aligned} L_0v=g(a,\bar{a},v,r) = -rk^2 w'' + k^2 (w^2)'', \end{aligned}$$

(A.2)

where g is analytic in its argument and $g(0,0,0,r)=0$ for all $r\in {\mathbb {R}}$. We define the projection operator $\Pi :L^2({\mathbb {T}})\rightarrow \text {ker}L_0$ as

$$\begin{aligned} \Pi f(z)=\widehat{f}(1)e^{iz}+\widehat{f}(-1)e^{-iz}. \end{aligned}$$

Since $\Pi v=0$, we may write (A.2) as

$$\begin{aligned} L_0v=(I-\Pi )g(a,\bar{a},v,r)\quad \text {and} \quad 0=\Pi g(a,\bar{a},v,r). \end{aligned}$$

(A.3)

Note that, for wavenumbers not satisfying the resonance condition (2.4), we can write

$$\begin{aligned} f(z)= L_{0_{|(I-\Pi )H^{4}({\mathbb {T}})}} \left( \sum _{n\ne \pm 1}\frac{\widehat{f}(n)}{(-c_0k^2n^2 +\beta k^4 n^4 + \gamma )}e^{inz} \right) . \end{aligned}$$

or

$$\begin{aligned} ( L_{0_{|(I-\Pi )H^{4}({\mathbb {T}})}})^{-1}f(z) =\sum _{n\ne \pm 1}\frac{\widehat{f}(n)}{(-c_0k^2n^2 +\beta k^4 n^4 + \gamma )}e^{inz}. \end{aligned}$$

Consequently, we can rewrite (A.3) as

$$\begin{aligned} v=L_0^{-1}(I-\Pi )g(a,\bar{a},v,r)\quad \text {and} \quad 0=\Pi g(a,\bar{a},v,r). \end{aligned}$$

(A.4)

Note that $( L_{0_{|(I-\Pi )H^{4}({\mathbb {T}})}})^{-1}$ depends analytically on its arguments.

In parallel with the proof in [8], it follows from the repeated use of implicit function theorem and symmetries of the equation that a unique solution

$$\begin{aligned} (v,r)=(V(a,\bar{a},r),r(|a|)) \end{aligned}$$

exists to equations in (A.4) in the vicinity of $(a,\bar{a},r)=(0,0,0)$, which depends analytically on its argument. Moreover,

$$\begin{aligned} w(z)=0+a\cos z+V(a,a,r(|a|))(z)\quad \text {and}\quad c=c_0+r(|a|) \end{aligned}$$

solve (A.1) for |a| sufficiently small. Since details are exactly similar, we omit it here.

Since w and c depend analytically on a for |a| sufficiently small and since c is even in a, we write that

$$\begin{aligned} w(k;a)(z)= 0+ a \cos z+a^2w_2(z)+a^3w_3(z)+a^4w_4(z)+O(a^5) \end{aligned}$$

and

$$\begin{aligned} c(k;a)=c_0(k)+a^2c_2+a^4c_4+O(a^6) \end{aligned}$$

as $a\rightarrow 0$, where $u_2$, $u_3$ and $u_4$ are even and $2\pi$-periodic in z. Substituting these into (A.1), we see that the coefficients of a both sides are equal. Equating the coefficients of $a^2$, we arrive at

$$\begin{aligned} \beta k^4 w_2''''+c_0k^2 w_2''+\gamma w_2=-2k^2\cos (2z) \end{aligned}$$

which has solution

$$\begin{aligned} w_2(z)=A_2\cos (2z):=\frac{2k^2}{3\gamma -12\beta k^4}\cos (2z). \end{aligned}$$

(A.5)

For $O(a^3)$, we get

$$\begin{aligned} \beta k^4 w_3''''+c_0k^2 w_3''+\gamma w_3 =k^2(c_2-A_2)\cos (z)-9k^2 A_2\cos (3z) \end{aligned}$$

which has solution

$$\begin{aligned} c_2=A_2 \end{aligned}$$

(A.6)

and

$$\begin{aligned} w_3(z)=A_3\cos (3z):=\frac{9k^2A_2}{8\gamma -72\beta k^4}\cos (3z). \end{aligned}$$

(A.7)

For $O(a^4)$, we get

$$\begin{aligned} c_4=3A_2A_3-2A_2^3 \end{aligned}$$

(A.8)

and

$$\begin{aligned} w_4(z)= & {} A_{42}\cos 2z+A_{44}\cos 4z:=\left( 2A_2A_3-2A_2^3 \right) \cos 2z\nonumber \\&\quad +\left( \dfrac{8k^2(A_2^2+2A_3)}{15\gamma -240\beta k^4} \right) \cos 4z. \end{aligned}$$

(A.9)

This completes the proof.